A Math game （DFS+剪枝）

数学游戏求解策略

最新推荐文章于 2021-10-22 14:12:05 发布

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本文介绍了一款数学游戏，玩家需从给定数字集中选择数字组合以达到特定目标值。文章详细阐述了解题思路，采用深度优先搜索算法，并通过逆序排序及累积和优化策略来提高效率。

A Math game

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 256000/128000KB (Java/Others)

Problem Description

Recently, Losanto find an interesting Math game. The rule is simple: Tell you a number H, and you can choose some numbers from a set {a[1],a[2],......,a[n]}.If the sum of the number you choose is H, then you win. Losanto just want to know whether he can win the game.

Input

There are several cases.
In each case, there are two numbers in the first line n (the size of the set) and H. The second line has n numbers {a[1],a[2],......,a[n]}.0<n<=40, 0<=H<10^9, 0<=a[i]<10^9,All the numbers are integers.

Output

If Losanto could win the game, output "Yes" in a line. Else output "No" in a line.

Sample Input

10 87
2 3 4 5 7 9 10 11 12 13
10 38
2 3 4 5 7 9 10 11 12 13

Sample Output

No
Yes

Source

第九届北京化工大学程序设计竞赛

题目链接：http://acdream.info/problem?pid=1726

题目大意：取一组数字中的任意一些数，看能否组成目标数。

解题思路：深搜，很简单能想到，避免T需要做小小的优化。即递减排序以及从后往前记录数字和。如果当前值加上后面所有未使用的数字还小于目标值，说明已无解，不需要再往下递归。

代码如下：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define ll long long 
using namespace std;
ll a[45],vis[45],sum[45];
int n,p;
ll k;
int cmp(ll u,ll v)
{
	return u>v;
}
void dfs(int s,ll cur)
{
	if(k==cur||p)
	{
		p=1;
		return ;
	}
	if(s>=n)
		return;
	for(int i=s;i<n;i++)
	{
		if(vis[i])
			continue;
		if(cur+sum[i]<k)        //如果当前值加上后面所有未使用的数字还小于目标值，说明已无解，不需要再往下递归
			return;
		if(cur+a[i]<=k)
		{
			vis[i]=1;
			dfs(i+1,cur+a[i]);
			if(p)
				return;
			vis[i]=0;
		}
	}
}
int main(void)
{
	while(scanf("%d%I64d",&n,&k)!=EOF)
	{
		p=0;
		memset(vis,0,sizeof(vis));
		memset(sum,0,sizeof(sum));
	    for(int i=0;i<n;i++)
	    	scanf("%I64d",&a[i]);
	    sort(a,a+n,cmp);
	    for(int i=n-1;i>=0;i--)
	    	sum[i]=sum[i+1]+a[i];   //从后往前记录数字和
	    dfs(0,0);
	    if(p)
	    	printf("Yes\n");
	    else
	    	printf("No\n");
	}
}